Aptitude & reasoning

Square Root & Cube Root

 

Step 1: First of all group the number in pairs of 2 starting from the right.

 

Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.

 

Step 3: To get the unit’s place digit of the square root

 

Remember the following

If number ends in	Unit’s place digit of the square root
1	1 or 9(10-1)
4	2 or 8(10-2)
9	3 or 7(10-3)
6	4or 6(10-4)
5	5
0	0

 

Lets see the logic behind this for a better understanding

We know,

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

 

Now, observe the unit’s place digit of all the squares.

Do you find anything common?

 

We notice that,

Unit’s place digit of both 1² and 9² is 1.

Unit’s place digit of both 2² and 8² is 4

Unit’s place digit of both 3² and 7² is 9

Unit’s place digit of both 4² and 6² is 6.

Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.

 

Remember,

If first pair of the original number > Result obtained on multiplication then  select the greater number  out of the two numbers as the unit’s place digit of the square root.

 

If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.

 

 

Let us consider an example to get a better understanding of the method

 

 

Example 1: √784=?

Step 1: We start by grouping the numbers in pairs of two from right as follows

7 84

 

Step 2: To get the ten’s place digit,

We find that nearest square to first group (7) is 4 and √4=2

Therefore ten’s place digit=2

 

Step 3: To get the unit’s place digit,

We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).

 

Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,

2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8

Ans:√784=28

 

 

 

Cube roots of perfect cubes

It may take two-three minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in one-two seconds using Vedic Mathematics.

We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.

 

Points to remember  for speedy  calculation of cube roots

1. To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.

13	=	1
23	=	8
33	=	27
43	=	64
53	=	125
63	=	216
73	=	343
83	=	512
93	=	729
103	=	1000

2. From the above cubes of 1 to 10, we need to remember an interesting property.

13 = 1	=>	If last digit of the perfect cube = 1, last digit of the cube root = 1
23 = 8	=>	If last digit of the perfect cube = 8, last digit of the cube root = 2
33 = 27	=>	If last digit of the perfect cube = 7, last digit of the cube root = 3
43 = 64	=>	If last digit of the perfect cube = 4, last digit of the cube root = 4
53 = 125	=>	If last digit of the perfect cube =5, last digit of the cube root = 5
63 = 216	=>	If last digit of the perfect cube = 6, last digit of the cube root = 6
73 = 343	=>	If last digit of the perfect cube = 3, last digit of the cube root = 7
83 = 512	=>	If last digit of the perfect cube = 2, last digit of the cube root = 8
93 = 729	=>	If last digit of the perfect cube = 9, last digit of the cube root = 9
103 = 1000	=>	If last digit of the perfect cube = 0, last digit of the cube root = 0

 

It’s very easy to remember the relations given above because

1	->	1	(Same numbers)
8	->	2	(10’s complement of 8 is 2 and 8+2 = 10)
7	->	3	(10’s complement of 7 is 3 and 7+3 = 10)
4	->	4	(Same numbers)
5	->	5	(Same numbers)
6	->	6	(Same numbers)
3	->	7	(10’s complement of 3 is 7 and 3+7 = 10)
2	->	8	(10’s complement of 2 is 8 and 2+8 = 10)
9	->	9	(Same numbers)
0	->	0	(Same numbers)

 

Also see
8 ->  2 and 2 ->  8
7 -> 3 and 3-> 7

 

 

 

 

 

Questions

Level-I

1.	The cube root of .000216 is:
	A. .6 B. .06 C. 77 D. 87

 

2.	What should come in place of both x in the equation x = 162 . 128 x
	A. 12 B. 14 C. 144 D. 196

 

3.	The least perfect square, which is divisible by each of 21, 36 and 66 is:
	A. 213444 B. 214344 C. 214434 D. 231444

 

4.	1.5625 = ?
	A. 1.05 B. 1.25 C. 1.45 D. 1.55

 

5.	If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?
	A. 13.41 B. 20.46 C. 21.66 D. 22.35
6.	If a = 0.1039, then the value of 4a2 – 4a + 1 + 3a is:
	A. 0.1039 B. 0.2078 C. 1.1039 D. 2.1039

 

7.	If x = 3 + 1 and y = 3 – 1 , then the value of (x2 + y2) is: 3 – 1 3 + 1
	A. 10 B. 13 C. 14 D. 15

 

8.	A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
	A. 57 B. 67 C. 77 D. 87

 

9.	The square root of (7 + 35) (7 – 35) is
	A. 5 B. 2 C. 4 D. 35

 

 

 

10.	If 5 = 2.236, then the value of 5 – 10 + 125 is equal to: 2 5
	A. 5.59 B. 7.826 C. 8.944 D. 10.062       Level-II
11.	625 x 14 x 11 is equal to: 11 25 196
	A. 5 B. 6 C. 8 D. 11

 

12.	0.0169 x ? = 1.3
	A. 10 B. 100 C. 1000 D. None of these

 

13.	3 – 1 2 simplifies to: 3
	A. 3 4 B. 4 3 C. 4 3 D. None of these

 

14.	How many two-digit numbers satisfy this property.: The last digit (unit’s digit) of the square of the two-digit number is 8 ?
	A. 1 B. 2 C. 3 D. None of these

 

15.	The square root of 64009 is:
	A. 253 B. 347 C. 363 D. 803

 

 

16. √29929 = ?

A. 173 B. 163 C. 196 D. 186

 

 

 

 

 

 

17. √106.09 = ?
A.	10.6
B.	10.5
C.	10.3
D.	10.2

 

18.  ?/√196 = 5
A.	76
B.	72
C.	70
D.	75

 

Answers

Level-I

 

Answer:1 Option B

 

Explanation:

(.000216)1/3	=		216		1/3
			106

 

=		6 x 6 x 6		1/3
		102 x 102 x 102

 

=	6
	102

 

=	6
	100

= 0.06

 

Answer:2 Option A

 

Explanation:

Let	x	=	162
	128		x

Then x² = 128 x 162

= 64 x 2 x 18 x 9

= 8² x 6² x 3²

= 8 x 6 x 3

= 144.

x = 144 = 12.

 

Answer:3 Option A

 

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2² x 3² x 7² x 11² = 213444

 

Answer:4 Option B

 

Explanation:

1|1.5625( 1.25

|1

|——-

22| 56

| 44

|——-

245| 1225

| 1225

|——-

|    X

|——-

1.5625 = 1.25.

 

 

Answer:5 Option D

 

Explanation:

35 + 125 = 17.88

35 + 25 x 5 = 17.88

35 + 55 = 17.88

85 = 17.88

5 = 2.235

80 + 65 = 16 x 5 + 65

= 45 + 65

= 105 = (10 x 2.235) = 22.35

 

 

 

Answer:6 Option C

 

Explanation:

4a² – 4a + 1 + 3a = (1)² + (2a)² – 2 x 1 x 2a + 3a

= (1 – 2a)² + 3a

= (1 – 2a) + 3a

= (1 + a)

= (1 + 0.1039)

= 1.1039

 

Answer:7 Option C

 

Explanation:

x =	(3 + 1)	x	(3 + 1)	=	(3 + 1)2	=	3 + 1 + 23	= 2 + 3.
	(3 – 1)		(3 + 1)		(3 – 1)		2

 

y =	(3 – 1)	x	(3 – 1)	=	(3 – 1)2	=	3 + 1 – 23	= 2 – 3.
	(3 + 1)		(3 – 1)		(3 – 1)		2

x² + y² = (2 + 3)² + (2 – 3)²

= 2(4 + 3)

= 14

 

Answer:8 Option C

 

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = 5929 = 77

 

 

Answer:9 Option B

 

Explanation:

(7 + 35)(7 – 35)

=

(7)2 – (35)2

= 49 – 45  = 4  = 2

 

 

Answer:10 Option B

 

Explanation:

5	–	10	+ 125	=	(5)2 – 20 + 25 x 55
2		5			25

 

=	5 – 20 + 50
	25

 

=	35	x	5
	25		5

 

=	355
	10

 

=	7 x 2.236
	2

 

= 7 x 1.118

 

= 7.826

 

 

Level-II

Answer:11 Option A

 

Explanation:

Given Expression =	25	x	14	x	11	= 5.
	11		5		14

 

 

 

Answer:12 Option B

 

Explanation:

Let 0.0169 x x = 1.3.

Then, 0.0169x = (1.3)² = 1.69

x =	1.69	= 100
	0.0169

 

 

 

Answer:13 Option C

 

Explanation:

	3 –	1		2	= (3)2 +		1		2	– 2 x 3 x	1
		3					3				3

 

= 3 +	1	– 2
	3

 

= 1 +	1
	3

 

=	4
	3

 

 

 

Answer:14 Option D

 

Explanation:

A number ending in 8 can never be a perfect square.

 

 

Answer:15 Option A

 

Explanation:

2 |64009( 253      |4      |———-45  |240      |225      |———-503| 1509      |  1509      |———-      |     X      |———-

64009 = 253.

 

 

Answer:16 Option A

 

Explanation:
√29929 = So, √29929 = 173

 

 

Answer:17 Option C

 

Answer:18 Option C

 

Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

1. x km/hr = (x×5)/18 m/s

 

2. y m/s = (y×18)/5 km/hr

 

3. Speed = distance/time, that is, s = d/t

 

4. velocity = displacement/time, that is, v = d/t

 

5. Time taken by a train x meters long to pass a pole or standing man or a post
   = Time taken by the train to travel x meters.

 

6. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

7. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

8. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

9. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

10. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

11. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres	B. 160 metres
C. 200 metres Answer : Option C	D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec	B. 99 s
C. 89 s	D. 80 s

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s	B. 12 s
C. 9.8 s	D. 8 s

 

Answer : Option A

Explanation :

Distance = 140+160 = 300 m

Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s

Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr	B. 69 km/hr
C. 74 km/hr	D. 61 km/hr

 

Answer : Option A

Explanation :

Let x is the length of the train and v is the speed

Time taken to move the post = 8 s

=> x/v = 8

=> x = 8v — (1)

Time taken to cross the platform 264 m long = 20 s

(x+264)/v = 20

=> x + 264 = 20v —(2)

Substituting equation 1 in equation 2, we get

8v +264 = 20v

=> v = 264/12 = 22 m/s

= 22×36/10 km/hr = 79.2 km/hr

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3	B. 2 :1
C. 4 : 3	D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m	B. 190 m
C. 210 m	D. 230 m

 

Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s

time = 9s

Total distance covered = 270 + x where x is the length of other train

(270+x)/9 = 500/9

=> 270+x = 500

=> x = 500-270 = 230 meter

7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m	B. 10 a.m.
C. 9.10 a.m.	D. 11 a.m.

 

Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42	B. 36
C. 28	D. 20

 

Answer : Option B

Explanation :

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v+v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10×36/10 km/hr = 36 km/hr

 

Level 2

1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m	B. 245 m
C. 235 m	D. 220 m

 

Answer : Option B

Explanation :

Assume the length of the bridge = x meter

Total distance covered = 130+x meter

total time taken = 30s

speed = Total distance covered /total time taken = (130+x)/30 m/s

=> 45 × (10/36) = (130+x)/30

=> 45 × 10 × 30 /36 = 130+x

=> 45 × 10 × 10 / 12 = 130+x

=> 15 × 10 × 10 / 4 = 130+x

=> 15 × 25 = 130+x = 375

=> x = 375-130 =245

2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.
A. 182 km/hr	B. 180 km/hr
C. 152 km/hr	D. 169 km/hr

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data	B. 3 : 1
C. 1 : 3	D. 3 : 2

 

Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively

length of train1 = 27x

length of train2 = 17y

Relative speed= x+ y

Time taken to cross each other = 23 s

=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)

=> 4x = 6y => x/y = 6/4 = 3/2

4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46	B. 36
C. 18	D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these	B. 280 meter
C. 240 meter	D. 200 meter

 

Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s

Length of the train = speed × time taken to cross the man = 15×20 = 300 m

Let the length of the platform = L

Time taken to cross the platform = (300+L)/15

=> (300+L)/15 = 36

=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter

6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m	B. 54 m
C. 50 m	D. 55 m

 

Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = (v-2) (10/36) — (1)

x/10 = (v-4) (10/36) — (2)

Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m	B. 360 m
C. 480 m	D. 400 m

 

Answer : Option D

Explanation :

Speed of train1 = 48 kmph

Let the length of train1 = 2x meter

Speed of train2 = 42 kmph

Length of train 2 = x meter (because it is half of train1’s length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s

Time = 12 s

Distance/time = speed => 3x/12 = 25

=> x = 25×12/3 = 100 meter

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y where y is the length of the platform

=> 200 + y = 45×40/3 = 600

=> y = 400 meter

8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m	B. 500 m
C. 260 m	D. 430 m

 

Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the tunnel

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s

Distance/time = speed

(800+x)/60 = 65/3 => 800+x = 20×65 = 1300

=> x = 1300 – 800 = 500 meter

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m	B. 2779 m
C. 1329 m	D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

 

This topic has many practical application in day to day life. In engineering stage it is used in surveying. The basic purpose is to find the unknown variables by observing the angle of the line of sight. This is done by using some the fact that in a right angled triangle the ratio of any two sides is a function of the angle between them. From exam point of view this is one of the more tough sections and tedious to some extent. So an aspirant must thoroughly solve all the questions given here.

Important Formulas

1. Trigonometric Basics

sinθ=oppositeside/hypotenuse=y/r

cosθ=adjacentside/hypotenuse=x/r

tanθ=oppositeside/adjacentside=y/x

cosecθ=hypotenuse/oppositeside=r/y

secθ=hypotenuse/adjacentside=r/x

cotθ=adjacentside/oppositeside=x/y

From Pythagorean theorem, x2+y2=r2 for the right angled triangle mentioned above

 

2. Basic Trigonometric Values

 

θ in degrees	θ in radians	sinθ	cosθ	tanθ
0°	0	0	1	0
30°	π/6	1/2	3/√2	1/√3
45°	π/4	1/√2	1/√2	1
60°	π/3	3/√2	1/2	√3
90°	π/2	1	0	Not defined

 

3. Trigonometric Formulas

Degrees to Radians and vice versa

360°=2π radian

 

Trigonometry – Quotient Formulas

tanθ=sinθ/cosθ

cotθ=cosθ/sinθ

 

Trigonometry – Reciprocal Formulas

cosecθ=1/sinθ

secθ=1/cosθ

cotθ=1/tanθ

 

Trigonometry – Pythagorean Formulas

sin2θ+cos2θ=1

sec2θ−tan2θ=1

cosec2θ−cot2θ=1

 

4. Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, angle of elevation is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight).

i.e., angle of elevation =  AOP

5. Angle of Depression

Suppose a man from a point O looks down at an object P, placed below the level of his eye. Then, angle of depression is the angle between the horizontal and the observer’s line of sight

i.e., angle of depression =  AOP

6. Angle Bisector Theorem

Consider a triangle ABC as shown above. Let the angle bisector of angle A intersect side BC at a point D. Then BD/DC=AB/AC

(Note that an angle bisector divides the angle into two angles with equal measures.
i.e., BAD = CAD in the above diagram)

7. Few Important Values to memorize

√2=1.414, √3=1.732, √5=2.236

 

Solved Examples

Level 1

1.The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 12.4 m away from the wall. The length of the ladder is:
A. 14.8 m	B. 6.2 m
C. 12.4 m	D. 24.8 m

 

 

Answer : Option D

Explanation :

Consider the diagram shown above where PR represents the ladder and RQ represents the wall.

cos 60° = PQ/PR

1/2=12.4/PR

PR=2×12.4=24.8 m

2.From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:
A. 346 m	B. 400 m
C. 312 m	D. 298 m

 

 

Answer : Option A

Explanation :

tan 30°=RQ/PQ

1/√3=200/PQ

PQ=200√3=200×1.73=346 m

3.The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
A. None of these	B. 60°
C. 45°	D. 30°

 

 

Answer : Option C

Explanation :

Consider the diagram shown above where QR represents the tree and PQ represents its shadow

We have, QR = PQ
Let QPR = θ

tan θ = QR/PQ=1 (since QR = PQ)

=> θ = 45°

i.e., required angle of elevation = 45°

4.An observer 2 m tall is 103√ m away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:
A. None of these	B. 12 m
C. 14 m	D. 10 m

 

 

Answer : Option B

Explanation :

SR = PQ = 2 m

PS = QR = 10√3m

tan 30°=TS/PS

1/3=TS/10√3

TS=10√3/√3=10 m

TR = TS + SR = 10 + 2 = 12 m

5.From a tower of 80 m high, the angle of depression of a bus is 30°. How far is the bus from the tower?
A. 40 m	B. 138.4 m
C. 46.24 m	D. 160 m

 

 

Answer : Option B

Explanation :

Let AC be the tower and B be the position of the bus.

Then BC = the distance of the bus from the foot of the tower.

Given that height of the tower, AC = 80 m and the angle of depression, DAB = 30°

ABC = DAB = 30° (Because DA || BC)

tan 30°=AC/BC=>tan 30°=80/BC=>BC = 80/tan 30°=80/(1/√3)=80×1.73=138.4 m

i.e., Distance of the bus from the foot of the tower = 138.4 m

6.Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 m long?
A. 30°	B. 60°
C. 45°	D. None of these

 

 

Answer : Option B

Explanation :

Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ = 6√3 m

Let the angle of elevation, RPQ = θ

From the right  PQR,

tanθ=RQ/PQ=18/6√3=3/√3=(3×√3)/( √3×√3)=3√3/3=√3

θ=tan−1(3√)=60°

 

Level 2

1.A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
A. 8 min 17 second	B. 10 min 57 second
C. 14 min 34 second	D. 12 min 23 second

 

 

Answer : Option B

Explanation :

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car

Then, ADC = 30° , ACB = 45°

Let AB = h, BC = x, CD = y

tan 45°=AB/BC=h/x

=>1=h/x=>h=x——(1)

tan 30°=AB/BD=AB/(BC + CD)=h/(x+y)

=>1/√3=h/(x+y)

=>x + y = √3h

=>y = √3h – x

=>y = √3h−h(∵ Substituted the value of x from equation 1 )

=>y = h(√3−1)

Given that distance y is covered in 8 minutes
i.e, distance h(√3−1) is covered in 8 minutes

Time to travel distance x
= Time to travel distance h (∵ Since x = h as per equation 1).

Let distance h is covered in t minutes

since distance is proportional to the time when the speed is constant, we have

h(√3−1)∝8—(A)

h∝t—(B)

(A)/(B)=>h(√3−1)/h=8/t

=>(√3−1)=8/t

=>t=8/(√3−1)=8/(1.73−1)=8/.73=800/73minutes ≈10 minutes 57 seconds

2.The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
A. 5 metres	B. 8 metres
C. 10 metres	D. 12 metres

 

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the tower and DE represents the pole

Given that AC = 15 m , ADB = 30°, AEC = 60°

Let DE = h

Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h), BD = CE

tan 60°=AC/CE=>√3=15/CE=>CE = 15√3— (1)

tan 30°=AB/BD=>1/√3=(15−h)/BD

=>1/√3=(15−h)/(15/√3)(∵ BD = CE and Substituted the value of CE from equation 1)

=>(15−h)=(1/√3)×(15/√3)=15/3=5

=>h=15−5=10 m

i.e., height of the electric pole = 10 m

 

3.Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
A. 300 m	B. 173 m
C. 273 m	D. 200 m

 

 

Answer : Option C

Explanation :

Let BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m,  BAD = 30° ,  BCD = 45°

tan 30° = BD/BA⇒1/√3=100/BA

⇒BA=100√3

tan 45° = BD/BC

⇒1=100/BC

⇒BC=100

Distance between the two ships = AC = BA + BC
=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273 m

4.From the top of a hill 100 m high, the angles of depression of the top and bottom of a pole are 30° and 60° respectively. What is the height of the pole?
A. 52 m	B. 50 m
C. 66.67 m	D. 33.33 m

 

Answer : Option C

Explanation :

Consider the diagram shown above. AC represents the hill and DE represents the pole

Given that AC = 100 m

XAD = ADB = 30° (∵ AX || BD )
XAE = AEC = 60° (∵ AX || CE)

Let DE = h

Then, BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h), BD = CE

tan 60°=AC/CE

=>√3=100/CE=>CE = 100/√3— (1)

tan 30°=AB/BD=>1/√3=(100−h)/BD

=>1/√3=(100−h)/(100/√3)(∵ BD = CE and Substituted the value of CE from equation 1 )

=>(100−h)=1/√3×100/√3=100/3=33.33=>h=100−33.33=66.67 m

i.e., the height of the pole = 66.67 m

5.A vertical tower stands on ground and is surmounted by a vertical flagpole of height 18 m. At a point on the ground, the angle of elevation of the bottom and the top of the flagpole are 30° and 60° respectively. What is the height of the tower?
A. 9 m	B. 10.40 m
C. 15.57 m	D. 12 m

 

 

Answer : Option A

Explanation :

Let DC be the vertical tower and AD be the vertical flagpole. Let B be the point of observation.

Given that AD = 18 m, ABC = 60°, DBC = 30°

Let DC be h.

tan 30°=DC/BC

1/√3=h/BC

h=BC√3—— (1)

tan 60°=AC/BC

√3=(18+h)/BC

18+h=BC×√3—— (2)

(1)/(2)=>h/(18+h)=(BC/√3)/(BC×√3)=1/3

=>3h=18+h=>2h=18=>h=9 m

i.e., the height of the tower = 9 m

6.A balloon leaves the earth at a point A and rises vertically at uniform speed. At the end of 2 minutes, John finds the angular elevation of the balloon as 60°. If the point at which John is standing is 150 m away from point A, what is the speed of the balloon?
A. 0.63 meter/sec	B. 2.16 meter/sec
C. 3.87 meter/sec	D. 0.72 meter/sec

 

 

Answer : Option B

Explanation :

Let C be the position of John. Let A be the position at which balloon leaves the earth and B be the position of the balloon after 2 minutes.

Given that CA = 150 m, BCA = 60°

tan 60°=BA/CA

√3=BA/150

BA=150√3

i.e, the distance travelled by the balloon = 150√3meters

time taken = 2 min = 2 × 60 = 120 seconds

Speed = Distance/Time=150√3/120=1.25√3=1.25×1.73=2.16 meter/second

7. The angles of depression and elevation of the top of a wall 11 m high from top and bottom of a tree are 60° and 30° respectively. What is the height of the tree?
A. 22 m	B. 44 m
C. 33 m	D. None of these

 

 

Answer : Option B

Explanation :

Let DC be the wall, AB be the tree.

Given that DBC = 30°, DAE = 60°, DC = 11 m

tan 30°=DC/BC

1/√3=11/BC

BC = 11√3 m

AE = BC =11√3 m—— (1)

tan 60°=ED/AE

√3=ED/11√3[∵ Substituted the value of AE from (1)]

ED =11√3×√3=11×3=33

Height of the tree = AB = EC = (ED + DC) = (33 + 11) = 44 m

 

8. Two vertical poles are 200 m apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find the heights of the poles.
A. 141 m and 282 m	B. 70.5 m and 141 m
C. 65 m and 130 m	D. 130 m and 260 m

 

 

Answer : Option B

Explanation :

Let AB and CD be the poles with heights h and 2h respectively

Given that distance between the poles, BD = 200 m

Let E be the middle point of BD.

Let AEB = θ and CED = (90-θ) (∵ given that angular elevations are complementary)

Since E is the middle point of BD, we have BE = ED = 100 m

From the right  ABE,
tanθ=AB/BE and tanθ=h/100

h = 100tanθ—— (1)

From the right  EDC,

tan(90−θ)=CD/ED

cotθ=2h/100[∵tan(90−θ)=cotθ]

2h =100cotθ—— (2)

(1) × (2) => 2h2=1002[∵tanθ×cotθ=tanθ×1/tanθ=1]

=>√2h=100

=>h=100/√2=(100×√2)/( √2×√2)=50√2=50×1.41=70.5

2h=2×70.5=141

i.e., the height of the poles are 70.5 m and 141 m.

9. To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and he is 5 m away from the wall, what is the length of the window?
A. 8.65 m	B. 2 m
C. 2.5 m	D. 3.65 m

 

 

Answer : Option D

Explanation :

Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the man and the wall, BE = 5 m,
DAF = 45° , CAF = 60°

From the diagram, AF = BE = 5 m

From the right  AFD, tan45°=DF/AF

1=DF/5

DF = 5—— (1)From the right  AFC, tan60°=CF/AF

√3=CF/5

CF=5√3—— (2)

Length of the window = CD = (CF – DF)

=5√3−5[∵ Substitued the value of CF and DF from (1) and (2)]=5(√3−1)=5(1.73−1)=5×0.73=3.65 m

10.The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°. What is the approximate height of the mountain?
A. 1.2 km	B. 0.6 km
C. 1.4 km	D. 2.7 km

 

 

Answer : Option D

Explanation :

Let A be the foot and C be the summit of a mountain.

Given that CAB = 45°

From the diagram, CB is the height of the mountain. Let CB = x

Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that DAY = 30°

It is also given that from the point D, the elevation is 60°

i.e., CDE = 60°

From the right  ABC,

tan45°=CB/AB

=>1=x/AB[∵ CB = x (the height of the mountain)]

=>AB = x—— (eq:1)

From the right  AYD,

sin30°=DY/AD

=>1/2=DY/2(∵ Given that AD = 2)

=> DY=1—— (eq:2)

cos30°=AY/AD=>√3/2=AY/2(∵ Given that AD = 2)=> AY=√3—— (eq:3)

From the right  CED, tan60°=CE/DE=>tan60°=(CB – EB)/YB∵ [CE = (CB – EB) and DE = YB)]

=>tan60°=(CB – DY)/(AB – AY)[ ∵ EB = DY and YB = (AB – AY)]

=>tan60°=(x – 1)/(x -√3)∵ [CB = x, DY = 1(eq:2), AB=x (eq:1) and AY = 3√(eq:3)]

=>√3=(x – 1)/(x -√3)=>x√3−3=x−1=>x(√3−1)=2=>0.73x=2=>x=2/0.73=2.7

i.e., the height of the mountain = 2.7 km

 

Classification involves putting things into a class or group according to particular characteristics so it’s easier to make sense of them, whether you’re organizing your shoes, your stock portfolio, or a group of invertebrates.  From all competitive examination classification is one of the most important topics, this pattern come with lot of questions minimum they asking the 4 to 5 question from the classification topic. In the SSC CGL or SSC constable GD examination having the same topics from the reasoning section but the standard of the topic will be different, so most of the candidates preference for this topic to get the best score in the written examination.

 

 

Directions: Find the odd one out

 

1. A. Square B. Circle                     C. Rectangle             D. Triangle

 

2. A. Cotton B. Terene                  C. Silk                         D. Wool

 

3. A. Light B. Wave                    C. Heat                      D. Sound

 

4. A. 81 : 243 B. 16 :64                   C. 64 : 192                D. 25 : 75

 

5. A. 64 : 8 B. 80 : 9                     C. 7 : 49                     D. 36 : 6

 

6. A. 26 : 62 B. 36 : 63                  C. 46 : 64                  D. 56 : 18

 

7. A. ABZY B. BCYX                      C. CDVW                   D. DEVU

 

8. A. ACE B. FHJ                         C. KLM                       D. SUW

 

9. Find the wrong number in the series

441, 484, 529, 566, 625

1. 484 B. 529                                    C. 625                                    D. 566

 

10. Find the wrong number in the series

232, 343, 454, 564, 676

1. 676 B. 454                                    C. 343                                    D. 564

 

 

SOLUTION TO CLASSIFICATION LEVEL 1

 

 

1. B. Except circle, all others are geometrical figures consisting straight lines.

 

2. B. Except terene, all others are natural fibres.

 

3. B. Except wave, all others are different form of energy.

 

4. B. 81*3=243

64*3=192

25*3=75

But     16*4=64

 

5. D. Except D, in each pair one number is square root of the other.

 

6. D. Except D, in each pair the position of digits has been interchanged.

 

7. C. A+1=B   &   Z-1=Y

B+1=C   &   Y-1=X

D+1=E   &   V-1=U

But   C+1=D   &   V+1=W

 

8. C. A+2=C    &   C+2=E

F+2=H     &   H+2=J

But      K+1=L     &   L+1=M

 

9. D. 21^2=441

22^2=484

23^2=529

25^2=625

But   (23.79)^2=566

 

10. D. 232+111=343

343+111=454

454+111=565 (but given 564)

Partnership : Partnership is an association of two or more parties, they put money for business.         Simple Partnership: Simple partnership is one in which the capitals of the partners are invested for the same time. The profit or losses are divided among the partners in the ratio of their investments.         Compound Partnership: Compound Partnership is one which the capitals of the partners are invested for different periods. In such cases equivalent capitals are calculated for a unit time by multiplying the capital with the number of units of time. The profits or losses are then divided in the ratio of these equivalent capitals. Tus the ratio of profits is directly proportional to both capital invested as time.         Working partner: A partner who participates in the working and manages the business is called a Working Partner.         Sleeping Partner: A partner who only invests capital but does not participate in the working of the business is called a Sleeping Partner.                 Division of Profit and Loss:     1. Rule :When investment of all partners are for the same time, the loss or profit is distributed among partners in the ratio of investment. Ex. Let P and Q invested Rs. a and b for one year in a business then share of profit and loss be , P’s share of profit : Q’s share profit = a : b 2.Rule : When investments are for different time period, then profit ratio is calculated as capital multiplied by length of investment Ex. P’s share of profit : Q’s share profit = a* t1 : b* t2     Questions with solutions Level-I

1. A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdrawns Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of 3 years ?

 

 

 

Solution: A : B : C = (40,000 X 36) : (80,000 X 12 + 40,000 X 24) : (120,000 X 24 + 40,000 X 12)   =     3: 4: 16

 

 

 

 

 

 

2. A, B, C enter into a partnership investing Rs. 35,000, Rs.45,000 and Rs.55,000 respectively. The respective shares of A, B, C in an annual profit of Rs.40,500 are ?

 

 

 

Solution : A : B : C = 35000 : 45000 : 55000 = 7 : 9 : 11.

 

A’s share = Rs (40500 x 7/27) = Rs. 10500

 

B’s share = Rs.(40500× 9/27) = Rs. 13500

 

C’s share = Rs.(40500×11/27)= Rs. 16500

 

 

 

 

 

 

 

3. In a business, Lucky invests Rs. 35,000 for 8 months and manju invests Rs 42,000 for 10 months. Out of a profit of Rs. 31,570. Manju’s share is 😕

 

 

Solution :      lucky: Manju = (35000 X 8) : (42,000 X 10) = 2:3
Manju’s share = Rs.3/5×31570 = Rs. 18,942

 

 

4. Amar started a business investing Rs. 70,000. Ramki joined him after six months with an amount of Rs. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business ?

 

 

Solution: Amar : Ramki : Sagar =

 

(70000 X 36) : (105000 X 30) : (140000 X24)   = 12 : 15 : 16.

 

 

 

5 . A begins a business with Rs 450 and is joined afterwards by B with  Rs 300. After how many months does B join if the profits at the end of the year is divided in the ratio 2 : 1?

 

 

Solution.-.(B) Suppose B joins for x months.

Then,     450 ´12    =    2
300 ´ x           1

x =450× 6
300

 

x= 9 months

\B joins after (12 – 9) = 3 months.

 

 

 

6. Shekhar started a business investing Rs. 25,000 in 1999. In 2000, he invested an additional amount of Rs. 10,000 and Rajeev joined him with an amount of Rs. 35,000. In 2001, Shekhar invested another additional amount of Rs. 10,000 and Jatin joined them with an amount of Rs. 35,000. What will be Rajeev’s share in the profit of Rs. 1,50,000 earned at the end of 3 years from the start of the business in 1999?.

 
Solution : Shekhar : Rajeev : Jatin  =

(25000  X  12 + 35000  X  12 + 45000  X  12) : (35000  X 24) :   (35000  X  12)
= 1260000   :  840000  :  420000  =   3  :  2  :  1.
Rajeev’s share   =  Rs.(150000×26)  =   Rs. 50000

 

 

 

 

 

7. A,B and C started a business with Rs.15000, Rs.25000 and Rs.35000 respectively.  A was paid 10% of the total profit as a salary and the balance was divided in the ration of investment.  If A’s share is Rs.4,200, then C’s share is: ?

 

 

 

Solution : A, B and C must divide their salaries in the ratio :

15,000 : 25,000:35,000 = 3:5:7
Assume total Profit = 100X.

then A share is 10% of 100X for managing business and 3/15 part of 90X for his investment (as the remaining profit is   (100X – 10X = 90X)
So total A’s share  =  10X  + 315 × 90X =  4,200
⇒X = 150
Substituting X  = 150 in 90X we get remaining profit for sharing. That is Rs.13,500
Now C’s share  = 715×13,500  =  Rs.6,300

 

 

Level-II

1.	A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855, the total profit is:
	A. Rs. 1425 B. Rs. 1500 C. Rs. 1537.50 D. Rs. 1576 Answer:1 Option B Explanation: Let the total profit be Rs. 100. After paying to charity, A’s share = Rs. 95 x 3 = Rs. 57. 5 If A’s share is Rs. 57, total profit = Rs. 100. If A’s share Rs. 855, total profit = 100 x 855 = 1500
2.	A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.
	A. Rs. 1900 B. Rs. 2660 C. Rs. 2800 D. Rs. 2840 Answer: 2 Option B   Explanation: For managing, A received = 5% of Rs. 7400 = Rs. 370. Balance = Rs. (7400 – 370) = Rs. 7030. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10  B’s share = Rs. 7030 x 14 = Rs. 2660. 37

	3 .A, B and C enter into a partnership in the ratio  :  : . After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:
	A. Rs. 2100 B. Rs. 2400 C. Rs. 3600 D.   Rs. 4000 Answer:3 Option D   Explanation: Ratio of initial investments = 7 : 4 : 6 = 105 : 40 : 36. 2 3 5 Let the initial investments be 105x, 40x and 36x.  A : B : C = 105x x 4 + 150 x 105x x 8 : (40x x 12) : (36x x 12) 100 = 1680x : 480x : 432x = 35 : 10 : 9. Hence, B’s share = Rs. 21600 x 10 = Rs. 4000. 54
4.	A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:
	A. Rs. 8400 B. Rs. 11,900 C. Rs. 13,600 D. Rs. 14,700   Answer:4 Option D   Explanation: Let C = x. Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000. So, x + x + 5000 + x + 9000 = 50000 3x = 36000 x = 12000 A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.  A’s share = Rs. 35000 x 21 = Rs. 14,700. 50

5.	Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?
	A. 5 : 7 : 8 B. 20 : 49 : 64 C. 38 : 28 : 21 D. None of these   Answer:5 Option B   Explanation: Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively. Then, 14x : 8y : 7z = 5 : 7 : 8. Now, 14x = 5        98x = 40y        y = 49 x 8y 7 20   And, 14x = 5        112x = 35z        z = 112 x = 16 x. 7z 8 35 5    x : y : z = x : 49 x : 16 x = 20 : 49 : 64. 20 5
6.	A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B’s contribution in the capital?
	A. Rs. 7500 B. Rs. 8000 C. Rs. 8500 D. Rs. 9000 Answer:6 Option D   Explanation: Let B’s capital be Rs. x. Then, 3500 x 12 = 2 7x 3 14x = 126000 x = 9000.

7.	A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew  of his capital and B withdrew  of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:
	A. Rs. 330 B. Rs. 360 C. Rs. 380 D. Rs. 430 Answer:7 Option A       Explanation: A : B = 4x x 3 + 4x – 1 x 4x x 7 : 5x x 3 + 5x – 1 x 5x x 7 4 5 = (12x + 21x) : (15x + 28x) = 33x :43x = 33 : 43.    A’s share = Rs. 760 x 33 = Rs. 330. 76

8.	A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
	A. 3 : 5 : 2 B. 3 : 5 : 5 C. 6 : 10 : 5 D. Data inadequate Answer:8 Option C   Explanation: Let the initial investments of A and B be 3x and 5x. A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.

9.	A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?
	A. Rs. 45 B. Rs. 50 C. Rs. 55 D. Rs. 60 Answer:9 Option A   Explanation: A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.  C’s rent = Rs. 175 x 9 = Rs. 45. 35

10.	A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B’s share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?
	A. Rs. 7500 B. Rs. 9000 C. Rs. 9500 D. Rs. 10,000 Answer: 10 Option A   Explanation: A : B : C = (20,000 x 24) : (15,000 x 24) : (20,000 x 18) = 4 : 3 : 3.  B’s share = Rs. 25000 x 3 = Rs. 7,500. 10

Surds

A surd is a square root which cannot be reduced to a rational number.

For example,  is not a surd.

However  is a surd.

If you use a calculator, you will see that  and we will need to round the answer correct to a few decimal places. This makes it less accurate.

If it is left as , then the answer has not been rounded, which keeps it exact.

Here are some general rules when simplifying expressions involving surds.

 

 

 

1. a^mx aⁿ = a^m + n

am	= am – n
an

• (a^m)ⁿ= a^mn

 

1. (ab)ⁿ= aⁿbⁿ

 

	a		n	=	an
	b				bn

1. a⁰= 1

 

 

Questions

Level-I

 

1.	(17)3.5 x (17)? = 178
	A. 2.29 B. 2.75 C. 4.25 D. 4.5

 

2.	If a x – 1 = b x – 3 , then the value of x is: b a
	A. 1 2 B. 1 C. 2 D. 7 2

 

3.	Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
	A. 1.45 B. 1.88 C. 2.9 D. 3.7

 

4.	If 5a = 3125, then the value of 5(a – 3) is:
	A. 25 B. 125 C. 625 D. 1625

 

5.	If 3(x – y) = 27 and 3(x + y) = 243, then x is equal to:
	A. 0 B. 2 C. 4 D. 6

 

.6.	(256)0.16 x (256)0.09 = ?
	A. 4 B. 16 C. 64 D. 256.25

 

7.	The value of [(10)150 ÷ (10)146]
	A. 1000 B. 10000 C. 100000 D. 106

 

8.	1  + 1 + 1 = ? 1 + x(b – a) + x(c – a) 1 + x(a – b) + x(c – b) 1 + x(b – c) + x(a – c)
	A. 0 B. 1 C. xa – b – c D. None of these

 

9.	(25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
	A. 8.5 B. 13 C. 16 D. 17.5 E. None of these

 

10.	(0.04)-1.5 = ?
	A. 25 B. 125 C. 250 D. 625

 

 

Level-II

 

11.	(243)n/5 x 32n + 1 = ? 9n x 3n – 1
	A. 1 B. 2 C. 9 D. 3n

 

12.	1 + 1 = ? 1 + a(n – m) 1 + a(m – n)
	A. 0 B. 1 2 C. 1 D. am + n

 

13.	If m and n are whole numbers such that mn = 121, the value of (m – 1)n + 1 is:
	A. 1 B. 10 C. 121 D. 1000

 

14.	xb (b + c – a) . xc (c + a – b) . xa (a + b – c) = ? xc xa xb
	A. xabc B. 1 C. xab + bc + ca D. xa + b + c

 

15. If 5√5 * 5³÷ 5^-3/2= 5^a+2 , the value of a is:
    A. 4
    B. 5
    C. 6
    D. 8

16.(132)7 ×(132)? =(132)11.5. A. 3 B. 3.5 C. 4 D. 4.5     17. (ab)x−2=(ba)x−7. What is the value   of x ?   A. 3 B. 4 C. 3.5 D. 4.5       18. (0.04)-2.5 = ?   A. 125 B. 25 C. 3125 D. 625

Answers

Level-I

Answer:1 Option D

 

Explanation:

Let (17)^3.5 x (17)^x = 17⁸.

Then, (17)^{3.5 + x} = 17⁸.

3.5 + x = 8

x = (8 – 3.5)

x = 4.5

 

Answer:2 Option C

 

Explanation:

Given	a		x – 1	=		b		x – 3
	b					a

 

a

x – 1

=

a

-(x – 3)

=

a

(3 – x)

b

b

b

x – 1 = 3 – x

2x = 4

x = 2.

 

 

Answer:3 Option C

 

Explanation:

x^z = y²        10^(0.48z) = 10^{(2 x 0.70)} = 10^1.40

0.48z = 1.40

z =	140	=	35	= 2.9 (approx.)
	48		12

 

Answer:4 Option A

 

Explanation:

5^a = 3125        5^a = 5⁵

a = 5.

5^{(a – 3)} = 5^{(5 – 3)} = 5² = 25.

 

 

Answer:5 Option C

 

Explanation:

3^x – y = 27 = 3³        x – y = 3 ….(i)

3^x + y = 243 = 3⁵        x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4

 

 

Answer:6 Option A

 

Explanation:

(256)^0.16 x (256)^0.09 = (256)^{(0.16 + 0.09)}

= (256)^0.25

= (256)^(25/100)

= (256)^(1/4)

= (4⁴)^(1/4)

= 4^4(1/4)

= 4¹

= 4

Answer:7 Option B

 

Explanation:

(10)150 ÷ (10)146 =	10150
	10146

= 10^{150 – 146}

= 10⁴

= 10000.

 

Answer:8 Option B

 

Explanation:

Given Exp. =	1  + 1  + 1 1 + xb + xc xa xa 1 + xa + xc xb xb 1 + xb + xa xc xc

 

=	xa	+	xb	+	xc
	(xa + xb + xc)		(xa + xb + xc)		(xa + xb + xc)

 

=	(xa + xb + xc)
	(xa + xb + xc)

= 1.

 

Answer:9 Option B

 

Explanation:

Let (25)^7.5 x (5)^2.5 ÷ (125)^1.5 = 5^x.

Then,	(52)7.5 x (5)2.5	= 5x
	(53)1.5

 

	5(2 x 7.5) x 52.5	= 5x
	5(3 x 1.5)

 

	515 x 52.5	= 5x
	54.5

5^x = 5^{(15 + 2.5 – 4.5)}

5^x = 5¹³

x = 13.

 

Answer:10 Option B

 

Explanation:

(0.04)-1.5 =		4		-1.5
		100

 

=		1		-(3/2)
		25

= (25)^(3/2)

= (5²)^(3/2)

= (5)^{2 x (3/2)}

= 5³

= 125.

 

Level-II

 

Answer:11 Option C

 

Explanation:

Given Expression	= (243)(n/5) x 32n + 1 9n x 3n – 1
	= (35)(n/5) x 32n + 1 (32)n x 3n – 1
	= (35 x (n/5) x 32n + 1) (32n x 3n – 1)
	= 3n x 32n + 1 32n x 3n – 1
	= 3(n + 2n + 1) 3(2n + n – 1)
	= 33n + 1 33n – 1
	= 3(3n + 1 – 3n + 1)   = 32   = 9.

Answer:12 Option C

 

Explanation:

1	+	1	=	1  + 1 1 + an am 1 + am an
1 + a(n – m)		1 + a(m – n)

 

=	am	+	an
	(am + an)		(am + an)

 

=	(am + an)
	(am + an)

= 1.

 

Answer:13 Option D

 

Explanation:

We know that 11² = 121.

Putting m = 11 and n = 2, we get:

(m – 1)^{n + 1} = (11 – 1)^{(2 + 1)} = 10³ = 1000.

 

Answer:14 Option B

 

Explanation:

Given Exp.	= x(b – c)(b + c – a) . x(c – a)(c +a – b) . x(a – b)(a + b – c)
	= x(b – c)(b + c) – a(b – c)  .  x(c – a)(c + a) – b(c – a) .  x(a – b)(a + b) – c(a – b)
	= x(b2 – c2 + c2 – a2 + a2 – b2)  .   x–a(b – c) – b(c – a) – c(a – b)
	= (x0 x x0)
	= (1 x 1) = 1.

 

Answer:15 option C

 

Answer:16

Explanation

am.an=am+n

(132)⁷ × (132)^x = (132)^11.5

=> 7 + x = 11.5

=> x = 11.5 – 7 = 4.5

 

 

Answer:17

Explanation:

an=1a−n

(ab)x−2=(ba)x−7⇒(ab)x−2=(ab)−(x−7)⇒x−2=−(x−7)⇒x−2=−x+7⇒x−2=−x+7⇒2x=9⇒x=92=4.5

 

Answer:18

Explanation:

a−n=1/an

(0.04)−2.5=(1/.04)2.5=(100/4)2.5=(25)2.5=(52)2.5=(52)(5/2)=55=3125

 

 

Introduction

Money is not free and it costs to borrow the money. Normally, the borrower has to pay an extra amount in addition to the amount he had borrowed. i.e, to repay the loan, the borrower has to pay the sum borrowed and the interest.

Lender and Borrower

The person giving the money is called the lender and the person taking the money is the borrower.

Principal (sum)

Principal (or the sum) is the money borrowed or lent out for a certain period. It is denoted by P.

Interest

Interest is the extra money paid by the borrower to the owner (lender) as a form of compensation for the use of the money borrowed.

Simple Interest (SI)

If the interest on a sum borrowed for certain period is calculated uniformly, it is called simple interest(SI).

Amount (A)

The total of the sum borrowed and the interest is called the amount and is denoted by A

• The statement “rate of interest 10% per annum” means that the interest for one year on a sum of Rs.100 is Rs.10. If not stated explicitly, rate of interest is assumed to be for one year.

 

• Let Principal = P, Rate = R% per annum and Time = T years. Then

  Simple Interest, SI = PRT/100

 

• From the above formula , we can derive the followings

  P=100×SI/RT

  R=100×SI/PT

  T=100×SI/PR

 

Some Formulae

1. If a sum of money becomes n times in T years at simple interest, then the rate of interest per annum can be given be R = 100(n−1)/T %
2. The annual instalment which will discharge a debt of D due in T years at R% simple interest per annum =100D/ (100T+RT(T-1)/2)
3. If an amount P₁is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by
   R=(P1R1+P2R2)/ (P1+P2)
4. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_nrespectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by (1/R1T1):(1/R2T2):⋯(1/RnTn)
5. If a certain sum of money P lent out for a certain time T amounts to P₁at R₁% per annum and to P₂at R₂% per annum, then P = (P2R1−P1R2)/ (R1−R2) and T = (P1−P2) ×100 years / (P2R1−P1R2)

SOLVED EXAMPLES

LEVEL 1

1.       Arun took a loan of Rs. 1400 with simple interest for as many years as the rate of interest. If he paid Rs.686 as interest at the end of the loan period, what was the rate of interest?
A. 8%	B. 6%
C. 4%	D. 7%

Ans. Let rate = R%

Then, Time, T = R years

P = Rs.1400

SI = Rs.686

SI= PRT/100⇒686 = 1400 × R × R/100⇒686=14 Rx R ⇒49=Rx R ⇒R=7

i.e.,Rate of Interest was 7%. (D)

2.       How much time will it take for an amount of Rs. 900 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
A. 2 years	B. 3 years
C. 1 year	D. 4 years

 

 

Ans. P = Rs.900

SI = Rs.81

T = ?

R = 4.5%

T= 100×SI/PR = 100×81/(900×4.5) = 2 years (A)

3.       A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :
A. Rs. 700	B. Rs. 690
C. Rs. 650	D. Rs. 698

 

 

Ans. Simple Interest (SI) for 1 year = 854-815 = 39

Simple Interest (SI) for 3 years = 39 × 3 = 117

Principal = 815 – 117 = Rs.698 (D)

 

4.       A sum fetched a total simple interest of Rs. 929.20 at the rate of 8 p.a. in 5 years. What is the sum?
A. Rs. 2323	B. Rs. 1223
C. Rs. 2563	D. Rs. 2353

 

Ans. SI = Rs.929.20

P = ?

T = 5 years

R = 8%

P = 100×SI/RT=100×929.20/(8×5) = Rs.2323 (A)

5.       What will be the ratio of simple interest earned by certain amount at the same rate of interest for 5 years and that for 15 years?
A. 3 : 2	B. 1 : 3
C. 2 : 3	D. 3 : 1

 
Solution 1
Let Principal = P

Rate of Interest = R%

Required Ratio = (PR×5/100)/ (PR×15/100) =1:3 (B)
Solution 2

Simple Interest = PRT100

Here Principal(P) and Rate of Interest (R) are constants

Hence, Simple Interest ∝ T

Required Ratio = Simple Interest for 5 years Simple Interest for 15 years=T1T2=515=13=1:3 (B)

6.       A sum of money amounts to Rs.9800 after 5 years and Rs.12005 after 8 years at the same rate of simple interest. The rate of interest per annum is
A. 15%	B. 12%
C. 8%	D. 5%

 

 

Ans. Simple Interest for 3 years = (Rs.12005 – Rs.9800) = Rs.2205

Simple Interest for 5 years = 22053×5=Rs.3675

Principal (P) = (Rs.9800 – Rs.3675) = Rs.6125

R = 100×SI/PT=100×3675/(6125×5) =12% (B)

7.       A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A. 5%	B. 10%
C. 7%	D. 8%

 

 

Ans. Let the rate of interest per annum be R%

Simple Interest for Rs. 5000 for 2 years at rate R% per annum +Simple Interest for Rs. 3000 for 4 years at rate R% per annum = Rs.2200

⇒5000×R×2/100+3000×R×4/100=2200

⇒100R + 120R=2200⇒220R=2200⇒R=10

i.e, Rate = 10%. (B)

8.       In how many years, Rs. 150 will produce the same interest at 6% as Rs. 800 produce in 2 years at 4½% ?
A. 4 years	B. 6 years
C. 8 years	D. 9 years

 

 

Ans. Let Simple Interest for Rs.150 at 6% for n years = Simple Interest for Rs.800 at 4½ % for 2 years

150×6×n/100=800×4.5×2/100

150×6×n=800×4.5×2

n=8 years (C)

 

LEVEL 2

1.        Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A. Rs. 6400	B. Rs. 7200
C. Rs. 6500	D. Rs. 7500

 

 

Ans. Let the investment in scheme A be Rs.x

and the investment in scheme B be Rs. (13900 – x)

We know that SI = PRT/100

Simple Interest for Rs.x in 2 years at 14% p.a. = x×14×2100=28x100Simple Interest for Rs.(13900 – x) in 2 years at 11% p.a. = (13900−x)×11×2/100 =22(13900−x)/100

Total interest =Rs.3508

Thus, 28x/100+22(13900−x)/100 = 3508

28x+305800−22x=350800

6x = 45000

x=45000/6=7500

Investment in scheme B = 13900 – 7500 = Rs.6400 (A)

2.       A certain sum in invested for T years. It amounts to Rs. 400 at 10% per annum. But when invested at 4% per annum, it amounts to Rs. 200. Find the time (T).
A. 45 years	B. 60 years
C. 40 years	D. 50 Years

 
Solution 1
Let the principal = Rs.x

and time = y years

Principal,x amounts to Rs.400 at 10% per annum in y years

Simple Interest = (400-x)

Simple Interest = PRT/100

⇒ (400−x) = x×10×y/100

⇒ (400−x) = xy/10— (equation 1)

Principal,x amounts to Rs.200 at 4% per annum in y years

Simple Interest = (200-x)

Simple Interest = PRT/100

⇒ (200−x) = x×4×y/100

⇒ (200−x) = xy/25— (equation 2)

(equation 1)/(equation2)

⇒(400−x) / (200−x) = (xy/10)/(xy/25)

⇒ (400−x)/ (200−x) =25/10

⇒ (400−x)/ (200−x) =52

⇒800−2x = 1000−5x

⇒200=3x

⇒x =200/3 Substituting this value of x in Equation 1, we get,

(400−200/3) = (200y/3)/10

⇒ (400−200/3) = 20y/3

⇒1200−200=20y

⇒1000=20y

y=1000/20=50 years (D)

Solution 2
If a certain sum of money P lent out for a certain time T amounts to P₁ at R₁% per annum and to P₂ at R₂% per annum, then

P = (P2R1−P1R2)/ (R1−R2)

T = (P1−P2)x 100 years/(P2R1−P1R2)

R₁ = 10%, R₂ = 4%

P₁ = 400, P₂ = 200

T = (P1−P2)x 100 / (P2R1−P1R2) = (400−200)x 100 / (200×10−400×4)

=200 x 100/ (2000−1600) =200 ×100/400 = 12×100=50 years (D)

3.       Mr. Mani invested an amount of Rs. 12000 at the simple interest rate of 10% per annum and another amount at the simple interest rate of 20% per annum. The total interest earned at the end of one year on the total amount invested became 14% per annum. Find the total amount invested.
A. Rs. 25000	B. Rs. 15000
C. Rs. 10000	D. Rs. 20000

 

Ans. If an amount P₁ is lent out at simple interest of R₁% per annum and another amount P₂ at simple interest rate of R₂% per annum, then the rate of interest for the whole sum can be given by

R= (P1R1+P2R2)/(P1+P2)

P₁ = Rs. 12000, R₁ = 10%

P₂ =? R₂ = 20%

R = 14%

14 = (12000×10+P2×20)/ (12000+P2)

12000×14+14P2 =120000+20P2

6P2=14×12000−120000=48000

⇒P2=8000

Total amount invested = (P₁ + P₂) = (12000 + 8000) = Rs. 20000 (D)

4.       A sum of money is lent at S.I. for 6 years. If the same amount is paid at 4% higher, Arun would have got Rs. 120 more. Find the principal
A. Rs. 200	B. Rs. 600
C. Rs. 400	D. Rs. 500

 

 

Ans. This means, simple interest at 4% for that principal is Rs.120

P=100×SI/ RT=100×120/ (4×6) =100×30/6 = 100×5 = 500 (D)

5.       The simple interest on Rs. 1820 from March 9, 2003 to May 21, 2003 at 7 1⁄2% rate is
A. Rs. 27.30	B. Rs. 22.50
C. Rs. 28.80	D. Rs. 29

 

 

Ans. Time, T = (22 + 30 + 21) days = 73 days = 73/365 year=1/5 year

Rate, R = 7.5%=15/2%

SI = PRT/100 = 1820× (15/2) × (1/5)/100 = 1820 × (3/2)/100 = 910 × 3/100

= 2730/100 = 27.30 (A)

6.       A sum of Rs. 7700 is to be divided among three brothers Vikas, Vijay and Viraj in such a way that simple interest on each part at 5% per annum after 1, 2 and 3 years respectively remains equal. The Share of Vikas is more than that of Viraj by
A. Rs.1200	B. Rs.1400
C. Rs.2200	D. Rs.2800

Ans. If a certain sum of money is lent out in n parts in such a manner that equal sum of money is obtained at simple interest on each part where interest rates are R₁, R₂, … , R_n respectively and time periods are T₁, T₂, … , T_n respectively, then the ratio in which the sum will be divided in n parts can be given by

1/R1T1:1/R2T2:⋯1/RnTn

 

T₁ = 1 , T₂ = 2, T₃ = 3

R₁ = 5 , R₂ = 5, R₃ = 5

Share of Vikas : Share of Vijay : Share of Viraj

= (1/5×1) : (1/5×2) : (1/5×3) = 1/1:1/2:1/3 = 6:3:2

Total amount is Rs. 7700

Share of Vikas = 7700×6/11=700×6 = 4200

Share of Viraj = 7700×2/11=700×2=1400

Share of Vikas is greater than Share of Viraj by (4200 – 1400) = Rs. 2800 (D)

 

7.       David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the total interest accrued in one year was Rs. 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
A. Rs.5000	B. Rs.2000
C. Rs.6000	D. Rs.3000

 

 

Ans. Let x, y and x be his investments in A, B and C respectively. Then

Then, Interest on x at 10% for 1 year

+ Interest on y at 12% for 1 year

+ Interest on z at 15% for 1 year

= 3200

x×10×1/100+y×12×1/100+z×15×1/100=3200

⇒10x+12y+15z=320000−−−(1)

Amount invested in Scheme C was 240% of the amount invested in Scheme B

=>z=240y/100 = 60y/25=12y/5−−−(2)

Amount invested in Scheme C was 150% of the amount invested in Scheme A

=>z=150x/100=3x/2

=>x=2z/3=2/3×12y/5=8y/5−−−(3)

From(1),(2) and (3),

10x + 12y + 15z = 320000

10(8y/5)+12y+15(12y/5)=320000

16y+12y+36y=320000

64y=320000

y=320000/64=10000/2=5000

i.e.,Amount invested in Scheme B = Rs.5000 (A)

 

 

In this module we will deal with basic concepts of time and distance, speed, average speed, conversion from km/h to m/s and vice versa. This chapter will form the basis of further concept of relative speed which is used in train and boat problems.

Important Formulas

1. Speed=Distance/Time
2. Distance=Speed×Time
3. Time=Distance/Speed
4. To convert Kilometers per Hour(km/hr) to Meters per Second(m/s)
   x km/hr=(x×5)/18m/s
5. To convert Meters per Second(m/s) to Kilometers per Hour(km/hr)
   x m/s=(x×18)/5 km/hr
6. If a car covers a certain distance at x kmph and an equal distance at y kmph, the average speed of the whole journey = 2xy/(x+y) kmph
7. Speed and time are inversely proportional (when distance is constant) ⇒Speed ∝1/Time (when distance is constant)
8. If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a

Solved Examples

Level 1

1.A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
A. 8.2	B. 4.2
C. 6.1	D. 7.2

 

Answer : Option D

Explanation :

Distance = 600 meter

time = 5 minutes = 5 x 60 seconds = 300 seconds

Speed = distance/time=600/300=2m/s=(2×18)/5 km/hr=36/5 km/hr=7.2 km/hr

2.Two boys starts from the same place walking at the rate of 5 kmph and 5.5 kmph respectively in the same direction. What time will they take to be 8.5 km apart?
A. 17 hr	B. 14 hr
C. 12 hr	D. 19 hr

 

Answer : Option A

Explanation :

Relative speed = 5.5 – 5 = .5 kmph (because they walk in the same direction)

distance = 8.5 km

Time = distance/speed=8.5/.5=17 hr.

3.Walking 6/7th of his usual speed, a man is 12 minutes too late. What is the usual time taken by him to cover that distance?
A. 1 hr 42 min	B. 1 hr
C. 2 hr	D. 1 hr 12 min

 

Answer : Option D

Explanation :

New speed = 6/7 of usual speed
Speed and time are inversely proportional.
Hence new time = 7/6 of usual time
Hence, 7/6 of usual time – usual time = 12 minutes
=>1/6 of usual time = 12 minutes => usual time = 12 x 6 = 72 minutes = 1 hour 12 minutes

4.A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
A. 3 km	B. 4 km
C. 5 km	D. 6 km

 

Answer : Option D

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph

Hence, average speed = (2×3×2)/(2+3)=12/5 km/hr .

Total time taken = 5 hours

⇒Distance travelled=(12/5)×5=12 km

⇒Distance between his house and office =12/2=6 km

5.If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. What is the actual distance travelled by him?
A. 80 km	B. 70 km
C. 60 km	D. 50 km

 

Answer : Option D

Explanation :

Assume that the person would have covered x km if travelled at 10 km/hr

⇒Speed = Distance/Time=x/10….. (Equation1)

Give that the person would have covered (x + 20) km if travelled at 14 km/hr
⇒Speed = Distance/Time=(x+20)/14….. (Equation2)

From Equations 1 and 2,
X/10=(x+20)/14⇒14x=10x+200⇒4x=200⇒x=200/4=50

6.A car travels at an average of 50 miles per hour for 212 hours and then travels at a speed of 70 miles per hour for 112 hours. How far did the car travel in the entire 4 hours?
A. 210 miles	B. 230 miles
C. 250 miles	D. 260 miles

 

Answer : Option B

Explanation :

Speed1 = 50 miles/hour

Time1 = 2*(1/2) hour=5/2 hour

⇒Distance1 = Speed1 × Time1 = (50×5)/2=25×5=125 miles

⇒Speed2 = 70 miles/hour

Time2 = 1*1/2 hour=3/2 hour

Distance2 = Speed2 × Time2 = 70×3/2=35×3=105 miles

Total Distance = Distance1 + Distance2 =125+105=230 miles

7.Sound is said to travel in air at about 1100 feet per second. A man hears the axe striking the tree, 11/5 seconds after he sees it strike the tree. How far is the man from the wood chopper?
A. 1800 ft	B. 2810 ft
C. 3020 ft	D. 2420 ft

 

Answer : Option D

Explanation :

Speed of the sound = 1100 ft/s ⇒Time = ¹¹/₅ second

Distance = Speed × Time = 1100 ×11/5=220×11=2420 ft

8.A man walking at the rate of 5 km/hr crosses a bridge in 15 minutes. What is the length of the bridge (in meters)?
A. 1250	B. 1280
C. 1320	D. 1340

 

Answer : Option A

Explanation :

Speed = 5 km/hr

Time = 15 minutes = 1/4 hour

Length of the bridge = Distance Travelled by the man

= Speed × Time = 5×1/4 km

=5×1/4×1000 metre=1250 metre

Level 2

1.A man takes 5 hours 45 min in walking to a certain place and riding back. He would have gained 2 hours by riding both ways. The time he would take to walk both ways is
A. 11 hrs	B. 8 hrs 45 min
C. 7 hrs 45 min	D. 9 hts 20 min

 

Answer : Option C

Explanation :

Given that time taken for riding both ways will be 2 hours lesser than the time needed for waking one way and riding back From this, we can understand that time needed for riding one way = time needed for waking one way – 2 hours Given that time taken in walking one way and riding back = 5 hours 45 min Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min In fact, you can do all these calculations mentally and save a lot of time which will be a real benefit for you. 2.A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
A. 121 km	B. 242 km
C. 224 km	D. 112 km

 

Answer : Option C

Explanation :

distance = speed x time
Let time taken to travel the first half = x hr
then time taken to travel the second half = (10 – x) hr
Distance covered in  the first half = 21x
Distance covered in  the second half = 24(10 – x)
But distance covered in  the first half = Distance covered in the second half
=> 21x = 24(10 – x) => 21x = 240 – 24x => 45x = 240 => 9x = 48 => 3x = 16⇒x=16/3

Hence Distance covered in the first half = 21x=21×16/3=7×16=112 km. Total distance = 2×112=224 km

3.A car traveling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. What is the actual speed of the car?
A. 30 km/hr	B. 35 km/hr
C. 25 km/hr	D. 40 km/hr

 

Answer : Option B

Explanation :

Time = 1 hr 40 min 48 sec = 1hr +40/60hr+48/3600hr=1+2/3+1/75=126/75hr

Distance = 42 kmSpeed=distance/time=42(126/75) =42×75/126

⇒5/7 of the actual speed = 42×75/126

⇒actual speed = 42×75/126×7/5=42×15/18=7×15/3=7×5=35 km/hr

4.A man covered a certain distance at some speed. If he had moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. What is the the distance in km?
A. 36	B. 38
C. 40	D. 42

 

Answer : Option C

Explanation :

Let the distance be x km , the speed in which he moved = v kmph

Time taken when moving at normal speed – time taken when moving 3 kmph faster = 40 minutes

⇒x/v−x/(v+3)=40/60⇒x[1/v−1/(v+3)]=2/3⇒x[(v+3−v)/v(v+3)]=2/3

⇒2v(v+3)=9x…………….(Equation1)

Time taken when moving 2 kmph slower – Time taken when moving at normal speed = 40 minutes
⇒x/(v−2)−x/v=40/60⇒x[1/(v−2)−1/v]=2/3

⇒x[(v−v+2)/v(v−2)]=2/3⇒x[2/v(v−2)]=2/3

⇒x[1/v(v−2)]=1/3⇒v(v−2)=3x…………….(Equation2)

Equation1/Equation2

⇒2(v+3)/(v−2)=3⇒2v+6=3v−6⇒v=12

Substituting this value of v inEquation1⇒2×12×15=9x

=>x= (2×12×15)/9= (2×4×15)/3=2×4×5=40. Hence distance = 40 km

5.In covering a distance of 30 km, Arun takes 2 hours more than Anil. If Arun doubles his speed, then he would take 1 hour less than Anil. What is Arun’s speed?
A. 8 kmph	B. 5 kmph
C. 4 kmph	D. 7 kmph

 

Answer : Option B

Explanation :

Let the speed of Arun = x kmph and the speed of Anil = y kmph
distance = 30 km

We know that distance/speed=time. Hence, 30/x−30/y=2………..(Equation1)

30/y−30/2x=1………..(Equation2)

Equation1 + Equation2⇒30/x−30/2x=3⇒30/2x=3⇒15/x=3⇒5/x=1⇒x=5. Hence Arun’s speed = 5 kmph

6.A car travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. What is the average speed for the first 320 km of the tour?
A. 70.24 km/hr	B. 74. 24 km/hr
C. 71.11 km/hr	D. 72.21 km/hr

 

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph,the average speed of the whole journey = 2xy/(x+y) kmph.

By using the same formula, we can find out the average speed quickly average speed = (2×64×80)/(64+80)=(2×64×80)/144⇒ (2×32×40)/36= (2×32×10)/9⇒ (64×10)/9=71.11 kmph

7.A man rides his bicycle 10 km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. What is his average speed for the entire trip approximately?
A. 11.2 kmph	B. 10 kmph
C. 10.2 kmph	D. 10.8 kmph

 

Answer : Option D

Explanation :

Total distance travelled = 10 + 12 = 22 km

Time taken to travel 10 km at an average speed of 12 km/hr = distance/speed=10/12 hr

Time taken to travel 12 km at an average speed of 10 km/hr = distance/speed=12/10 hr

Total time taken =10/12+12/10 hr

Average speed = distance/time=22/(10/12+12/10)=(22×120)/{(10×10)+(12×12)}

(22×120)/244=(11×120)/122=(11×60)/61=660/61≈10.8 kmph

8.An airplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 123 hours, it must travel at a speed of:
A. 660 km/hr	B. 680 km/hr
C. 700 km/hr	D. 720 km/hr

 

Answer : Option D

Explanation :

Speed and time are inversely proportional ⇒Speed ∝ 1/Time (when distance is constant)

Here distance is constant and Speed and time are inversely proportional

Speed ∝ 1/Time⇒Speed1/Speed2=Time2/Time1

⇒240/Speed2=(1*2/3)5⇒240/Speed2=(5/3)/5⇒240/Speed2=1/3⇒Speed2=240×3=720 km/hr

9.A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. What is the speed of the car?
A. 80 kmph	B. 102 kmph
C. 120 kmph	D. 140 kmph

 

Answer : Option C

Explanation :

Let speed of the car = x kmph

Then speed of the train = x *(100+50)/100=150 x /100=3 x /2 kmph

Time taken by the car to travel from A to B=75/x hours

Time taken by the train to travel from A to B=75/(3 x /2)+12.5/60 hours

Since both start from A at the same time and reach point B at the same time

75/x=75/(3 x /2)+12.5/60⇒25/x=12.5/60⇒x=(25×60)/12.5=2×60=120

TIME AND WORK

In these problems the number of persons, quantity of work done and time taken are important factors. Also time taken by a person depends on the efficiency of that person which comes into picture when different people do the work such as women, children do the work alongside the men. The problems related to time and work can be solved by two major approaches – ratio & proportions and unitary method. Let us proceed to find some formulae related to these questions.

Important Formulas – Time and Work

• If A can do a piece of work in n days, work done by A in 1 day = 1/n

 

• If A does 1/n work in a day, A can finish the work in n days

 

• If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then

M1 D1 H1 / W1 = M2 D2 H2 / W2

 

• If A can do a piece of work in p days and B can do the same in q days, A and B together can finish it in pq / (p+q) days

 

• If A is thrice as good as B in work, then

Ratio of work done by A and B = 3:1

Ratio of time taken to finish a work by A and B = 1: 3

 

SOLVED EXAMPLES

Level 1

1.P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
A. 8/15	B. 7/15
C. 11/15	D. 2/11

 

Answer : Option A

Explanation :

Amount of work P can do in 1 day = 1/15

Amount of work Q can do in 1 day = 1/20

Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60

Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15

Fraction of work left = 1 – 7/15= 8/15

2.A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. B alone can complete the work in — hours.
A. 12 hours	B. 6 hours
C. 8 hours	D. 10 hours

 

Answer : Option A

Explanation :

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = 1/4+1/3 = 7/12

Work done by B in 1 hour = 7/12 – 1/2 = 1/12

=> B alone can complete the work in 12 hours

3.A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?
A. 37 ½ days	B. 22 days
C. 31 days	D. 22 days

 

Answer : Option A

Explanation :

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 — (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 —(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 ½ days

4.P can finish a work in 18 days. Q can finish the same work in 15 days. Q worked for 10 days and left the job. How many days does P alone need to finish the remaining work?
A. 8	B. 5
C. 4	D. 6

 

Answer : Option D

Explanation :

Work done by P in 1 day = 1/18

Work done by Q in 1 day = 1/15

Work done by Q in 10 days = 10/15 = 2/3

Remaining work = 1 – 2/3 = 1/3

Number of days in which P can finish the remaining work = (1/3) / (1/18) = 6

5.Anil and Suresh are working on a special assignment. Anil needs 6 hours to type 32 pages on a computer and Suresh needs 5 hours to type 40 pages. If both of them work together on two different computers, how much time is needed to type an assignment of 110 pages?
A. 7 hour 15 minutes	B. 7 hour 30 minutes
C. 8 hour 15 minutes	D. 8 hour 30 minutes

 

Answer : Option C

Explanation :

Pages typed by Anil in 1 hour = 32/6 = 16/3

Pages typed by Suresh in 1 hour = 40/5 = 8

Pages typed by Anil and Suresh in 1 hour = 16/3 + 8 = 40/3

Time taken to type 110 pages when Anil and Suresh work together = 110 × 3 /40 = 33/4

= 8 ¼ hours = 8 hour 15 minutes

6.P works twice as fast as Q. If Q alone can complete a work in 12 days, P and Q can finish the work in — days
A. 1	B. 2
C. 3	D. 4

 

Answer : Option D

Explanation :

Work done by Q in 1 day = 1/12

Work done by P in 1 day = 2 × (1/12) = 1/6

Work done by P and Q in 1 day = 1/12 + 1/6 = ¼

=> P and Q can finish the work in 4 days

7.A work can be finished in 16 days by twenty women. The same work can be finished in fifteen days by sixteen men. The ratio between the capacity of a man and a woman is
A. 1:3	B. 4:3
C. 2:3	D. 2:1

 

Answer : Option B

Explanation :

Work done by 20 women in 1 day = 1/16 Work done by 1 woman in 1 day = 1/(16×20) Work done by 16 men in 1 day = 1/15 Work done by 1 man in 1 day = 1/(15×16) 8.P,Q and R together earn Rs.1620 in 9 days. P and R can earn Rs.600 in 5 days. Q and R in 7 days can earn Rs.910. How much amount does R can earn per day?
A. Rs.40	B. Rs.70
C. Rs.90	D. Rs.100

 

Answer : Option B

Explanation :

Amount Earned by P,Q and R in 1 day = 1620/9 = 180 —(1)

Amount Earned by P and R in 1 day = 600/5 = 120 —(2)

Amount Earned by Q and R in 1 day = 910/7 = 130 —(3)

(2)+(3)-(1) => Amount Earned by P , Q and 2R in 1 day

– Amount Earned by P,Q and R in 1 day = 120+130-180 = 70

=>Amount Earned by R in 1 day = 70
Ratio of the capacity of a man and woman =1/(15×16) : 1/(16×20) = 1/15 : 1/20

= 1/3 :1/4 = 4:3

Level 2

1.P, Q and R can do a work in 20, 30 and 60 days respectively. How many days does it need to complete the work if P does the work and he is assisted by Q and R on every third day?
A. 10 days	B. 14 days
C. 15 days	D. 9 days

 

Answer : Option C

Explanation :

Amount of work P can do in 1 day = 1/20

Amount of work Q can do in 1 day = 1/30

Amount of work R can do in 1 day = 1/60

P is working alone and every third day Q and R is helping him

Work completed in every three days = 2 × (1/20) + (1/20 + 1/30 + 1/60) = 1/5

So work completed in 15 days = 5 × 1/5 = 1

Ie, the work will be done in 15 days

2.A is thrice as good as B in work. A is able to finish a job in 60 days less than B. They can finish the work in – days if they work together.
A. 18 days	B. 22 ½ days
C. 24 days	D. 26 days

 

Answer : Option B

Explanation :

If A completes a work in 1 day, B completes the same work in 3 days

Hence, if the difference is 2 days, B can complete the work in 3 days

=> if the difference is 60 days, B can complete the work in 90 days

=> Amount of work B can do in 1 day= 1/90

Amount of work A can do in 1 day = 3 × (1/90) = 1/30

Amount of work A and B can together do in 1 day = 1/90 + 1/30 = 4/90 = 2/45

=> A and B together can do the work in 45/2 days = 22 ½ days

3.P can do a work in the same time in which Q and R together can do it. If P and Q work together, the work can be completed in 10 days. R alone needs 50 days to complete the same work. then Q alone can do it in A. 30 days B. 25 days C. 20 days D. 15 days   Answer : Option B Explanation : Work done by P and Q in 1 day = 1/10 Work done by R in 1 day = 1/50 Work done by P, Q and R in 1 day = 1/10 + 1/50 = 6/50 But Work done by P in 1 day = Work done by Q and R in 1 day . Hence the above equation can be written as Work done by P in 1 day × 2 = 6/50 => Work done by P in 1 day = 3/50 => Work done by Q and R in 1 day = 3/50 Hence work done by Q in 1 day = 3/50 – 1/50 = 2/50 = 1/25 So Q alone can do the work in 25 days 4.6 men and 8 women can complete a work in 10 days. 26 men and 48 women can finish the same work in 2 days. 15 men and 20 women can do the same work in – days.
A. 4 days	B. 6 days
C. 2 days	D. 8 days

 

Answer : Option A

Explanation :

Let work done by 1 man in 1 day = m and work done by 1 woman in 1 day = b

Work done by 6 men and 8 women in 1 day = 1/10

=> 6m + 8b = 1/10

=> 60m + 80b = 1 — (1)

Work done by 26 men and 48 women in 1 day = 1/2

=> 26m + 48b = ½

=> 52m + 96b = 1— (2)

Solving equation 1 and equation 2. We get m = 1/100 and b = 1/200

Work done by 15 men and 20 women in 1 day

= 15/100 + 20/200 =1/4

=> Time taken by 15 men and 20 women in doing the work = 4 days

5.Machine P can print one lakh books in 8 hours. Machine Q can print the same number of books in 10 hours while machine R can print the same in 12 hours. All the machines started printing at 9 A.M. Machine P is stopped at 11 A.M. and the remaining two machines complete work. Approximately at what time will the printing of one lakh books be completed?
A. 3 pm	B. 2 pm
C. 1:00 pm	D. 11 am

 

Answer : Option C

Explanation :

Work done by P in 1 hour = 1/8

Work done by Q in 1 hour = 1/10

Work done by R in 1 hour = 1/12

Work done by P,Q and R in 1 hour = 1/8 + 1/10 + 1/12 = 37/120

Work done by Q and R in 1 hour = 1/10 + 1/12 = 22/120 = 11/60

From 9 am to 11 am, all the machines were operating.

Ie, they all operated for 2 hours and work completed = 2 × (37/120) = 37/60.

6.A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in — days.
A. 5 5⁄11	B. 4 5⁄11
C. 6 4⁄11	D. 6 5⁄11

 

Answer : Option A

Explanation :

A can complete the work in 12 days working 8 hours a day

=> Number of hours A can complete the work = 12×8 = 96 hours

=> Work done by A in 1 hour = 1/96

B can complete the work in 8 days working 10 hours a day

=> Number of hours B can complete the work = 8×10 = 80 hours => Work done by B in 1 hour = 1/80

Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480 => A and B can complete the work in 480/11 hours. A and B works 8 hours a day.

Hence total days to complete the work with A and B working together = (480/11)/ (8) = 60/11 days = 5 ⁵⁄₁₁ days

Pending work = 1- 37/60 = 23/60

Hours taken by Q an R to complete the pending work = (23/60) / (11/60) = 23/11

which is approximately equal to 2. Hence the work will be completed approximately 2 hours after 11 am ; ie around 1 pm

7.If daily wages of a man is double to that of a woman, how many men should work for 25 days to earn Rs.14400? Given that wages for 40 women for 30 days are Rs.21600.
A. 12	B. 14
C. 16	D. 18

 

Answer : Option C

Explanation :

Wages of 1 woman for 1 day = 21600/(40×30)

Wages of 1 man for 1 day = (21600×2)/(40×30)

Wages of 1 man for 25 days = (21600×2×25)/(40×30)

Number of men = 14400/(21600×2×25)/(40×30)=144/(216×50)/40×30)=144/9=16

8.There is a group of persons each of whom can complete a piece of work in 16 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one more person joins them and this process continues till the work is completed. How many days are needed to complete the work?
A. 3 1⁄4 days	B. 4 1⁄3 days
C. 5 1⁄6 days	D. 6 1⁄5 days

 

Answer : Option C

Explanation :

Work completed in 1st day = 1/16

Work completed in 2nd day = (1/16) + (1/16) = 2/16

Work completed in 3rd day = (1/16) + (1/16) + (1/16) = 3/16

An easy way to attack such problems is from the choices. You can see the choices are

very close to each other. So just see one by one.

For instance, The first choice given in 3 ¹⁄₄

The work done in 3 days = 1/16 + 2/16 + 3/16 = (1+2+3)/16 = 6/16

The work done in 4 days = (1+2+3+4)/16 = 10/16

The work done in 5 days = (1+2+3+4+5)/16 = 15/16, almost close, isn’t it?

The work done in 6 days = (1+2+3+4+5+6)/16 > 1

Hence the answer is less than 6, but greater than 5. Hence the answer is 5 ¹⁄₆ days.

(Just for your reference, work done in 5 days = 15/16)

Pending work in 6th day = 1 – 15/16 = 1/16.

In 6th day, 6 people are working and work done = 6/16.

To complete the work 1/16, time required = (1/16) / (6/16) = 1/6 days.

Hence total time required = 5 + 1/6 = 5 ¹⁄₆ days

 

 

 

Introduction:

There are four main directions – East, West, North and South as shown below:

 

 

 

 

There are four cardinal directions – North-East (N-E), North-West (N-W), South-East (S-E), and South-West (S-W) as shown below:

 

 

 

Key points

 

1. At the time of sunrise if a man stands facing the east, his shadow will be towards west.
2. At the time of sunset the shadow of an object is always in the east.
3. If a man stands facing the North, at the time of sunrise his shadow will be towards his left and at the time of sunset it will be towards his right.
4. At 12:00 noon, the rays of the sun are vertically downward hence there will be no shadow

 

 

 

 

 

 

 

 

 

Practice Questions

Type 1:

Siva starting from his house, goes 5 km in the East, then he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution:

From third position it is clear he is 4 km from his house and is in North direction.

 

 

 

 

 

 

Type 2:

Suresh starting from his house, goes 4 km in the East, then he turns to his right and goes 3 km. What minimum distance will be covered by him to come back to his house?

Solution:

 

Type 3:

One morning after sunrise Juhi while going to school met Lalli at Boring road crossing. Lalli’s shadow was exactly to the right of Juhi. If they were face to face, which direction was Juhi facing?

 

Solution: In the morning sunrises in the east.

So in morning the shadow falls towards the west.

Now Lalli’s shadow falls to the right of the Juhi. Hence Juhi is facing South.

 

 

 

 

Type 4: Hema starting from her house walked 5 km to reach the crossing of Palace. In which direction she was going, a road opposite to this direction goes to Hospital. The road to the right goes to station. If the road which goes to station is just opposite to the road which IT-Park, then in which direction to Hema is the road which goes to IT-Park?

Solution:

From II it is clear that the road which goes to IT-Park is left to Hema.

 

 

 

 

 

 

 

 

Questions

 

Level-1

 

1.	One morning Udai and Vishal were talking to each other face to face at a crossing. If Vishal’s shadow was exactly to the left of Udai, which direction was Udai facing?
	A. East B. West C. North D. South

2.	Y is in the East of X which is in the North of Z. If P is in the South of Z, then in which direction of Y, is P?
	A. North B. South C. South-East D. None of these

3.	If South-East becomes North, North-East becomes West and so on. What will West become?
	A. North-East B. North-West C. South-East D. South-West

4.	A man walks 5 km toward south and then turns to the right. After walking 3 km he turns to the left and walks 5 km. Now in which direction is he from the starting place?
	A. West B. South C. North-East D. South-West

5.	Rahul put his timepiece on the table in such a way that at 6 P.M. hour hand points to North. In which direction the minute hand will point at 9.15 P.M. ?
	A. South-East   B. South   C. North D. West
6.	Rasik walked 20 m towards north. Then he turned right and walks 30 m. Then he turns right and walks 35 m. Then he turns left and walks 15 m. Finally he turns left and walks 15 m. In which direction and how many metres is he from the starting position?
	A. 15 m West B. 30 m East C. 30 m West D. 45 m East

7.

Two cars start from the opposite places of a main road, 150 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?

A. 65 km B. 75 km C. 80 km D. 85 km

8.	Starting from the point X, Jayant walked 15 m towards west. He turned left and walked 20 m. He then turned left and walked 15 m. After this he turned to his right and walked 12 m. How far and in which directions is now Jayant from X?
	A. 32 m, South B. 47 m, East C. 42 m, North D. 27 m, South

 

9.	One evening before sunset Rekha and Hema were talking to each other face to face. If Hema’s shadow was exactly to the right of Hema, which direction was Rekha facing?
	A. North B. South C. East D. Data is inadequate

10.	A boy rode his bicycle Northward, then turned left and rode 1 km and again turned left and rode 2 km. He found himself 1 km west of his starting point. How far did he ride northward initially?
	A. 1 km B. 2 km C. 3 km D. 5 km

 

 

Answers:

1Answer: Option C

Explanation:

 

2Answer: Option D

Explanation:

P is in South-West of Y.

 

3Answer: Option C

Explanation:

It is clear from the diagrams that new name of West will become South-East.

 

4Answer: Option D

Explanation:

Hence required direction is South-West.

 

5Answer: Option D

Explanation:

At 9.15 P.M., the minute hand will point towards west.

 

6Answer: Option D

Explanation:

 

7Answer: Option A

Explanation:

 

 

 

 

8Answer: Option A

Explanation:

 

9Answer: Option B

Explanation:

In the evening sun sets in West. Hence then any shadow falls in the East. Since Hema’s shadow was to the right of Hema. Hence Rekha was facing towards South.

 

10Answer: Option B

Explanation:

The boy rode 2 km. Northward

 

 

Level – 2

 

Dev, Kumar, Nilesh, Ankur and Pintu are standing facing to the North in a playground such as given below:

1. Kumar is at 40 m to the right of Ankur.
2. Dev is are 60 m in the south of Kumar.
3. Nilesh is at a distance of 25 m in the west of Ankur.
4. Pintu is at a distance of 90 m in the North of Dev

 

 

1.	Which one is in the North-East of the person who is to the left of Kumar?
	A. Dev B. Nilesh C. Ankur D. Pintu

2.	If a boy starting from Nilesh, met to Ankur and then to Kumar and after this he to Dev and then to Pintu and whole the time he walked in a straight line, then how much total distance did he cover?
	A. 215 m B. 155 m C. 245 m D.   185 m

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. Six flats on a floor in two rows facing North and South are allotted to P, Q, R, S, T and U.
2. Q gets a North facing flat and is not next to S.
3. S and U get diagonally opposite flats.
4. R next to U, gets a south facing flat and T gets North facing flat.

 

 

3.	If the flats of P and T are interchanged then whose flat will be next to that of U?
	A. P B. Q C. R D. T

4.	Which of the following combination get south facing flats?
	A. QTS B. UPT C. URP D. Data is inadequate

5.	The flats of which of the other pair than SU, is diagonally opposite to each other?
	A. QP B. QR C. PT D. TS
6.	Whose flat is between Q and S?
	A. T B. U C. R D. P

 

 

Directions to Solve

Each of the following questions is based on the following information:

1. 8-trees → mango, guava, papaya, pomegranate, lemon, banana, raspberry and apple are in two rows 4 in each facing North and South.
2. Lemon is between mango and apple but just opposite to guava.
3. Banana is at one end of a line and is just next in the right of guava or either banana tree is just after guava tree.
4. Raspberry tree which at one end of a line, is just diagonally opposite to mango tree.

 

 

	7 .Which of the following statements is definitely true?
	A. Papaya tree is just near to apple tree. B. Apple tree is just next to lemon tree. C. Raspberry tree is either left to Pomegranate or after. D. Pomegranate tree is diagonally opposite to banana tree.

8	Which tree is just opposite to raspberry tree?
	A. Papaya B. Pomegranate C. Papaya or Pomegranate D. Data is inadequate
9	Which tree is just opposite to banana tree?
	A. Mango B. Pomegranate C. Papaya D. Data is inadequate

 

 

Answer: 1 Option D

Explanation:

Ankur is in the left of Kumar. Hence Pintu is in North-East of Ankur

 

 

Answer: 2 Option A

Explanation:

Required distance = 25 m + 40 m + 60 m + 90 m

Required distance = 215 m

 

 

Answer:3 Option C

Explanation:

Hence flat R will be next to U.

 

 

Answer:4 Option C

 

Explanation:

Hence URP flat combination get south facing flats.

 

Answer:5 Option A

 

Explanation:

Hence QP is diagonally opposite to each other.

 

 

 

 

 

 

Answer:6 Option A

 

Explanation:

Hence flat T is between Q and S.

 

Answer: 7 Option B

 

Explanation:

 

 

Answer:8 Option C

 

Explanation:

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

Fractions
Any unit can be divided into any numbers of equal parts, one or more of this parts is called fraction of that unit. e.g. one-forth (1/4), one-third (1/3), three-seventh (3/7) etc.

The lower part indicates the number of equal parts into which the unit is divided, is called denominator. The upper part, which indicates the number of parts taken from the fraction is called the numerator. The numerator and the denominator of a fraction are called its terms.

• A fraction is unity, when its numerator and denominator are equal.
• A fraction is equal to zero if its numerator is zero.
• The denominator of a fraction can never be zero.
• The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number.e.g. 2/3 = 2/6 = 8/12 = (2/4)/(3/4)
• When there is no common factor between numerator and denominator it is called in its lowest terms.e.g. 15/25 = 3/5
• When a fraction is reduced to its lowest term, its numerator and denominator are prime to each other.
• When the numerator and denominator are divided by its HCF, fraction reduces to its lowest term.

Proper fraction: A fraction in which numerator is less than the denominator. e.g. 1/4, 3/4, 11/12 etc.

 

Improper Fraction:  A fraction in which numerator is equal to or more than the denominator. e.g. 5/4, 7/4, 13/12 etc.

 

Like fraction: Fractions in which denominators are same is called like fractions.

e.g. 1/12, 5/12, 7/12, 13/12 etc.

 

Unlike fraction: Fractions in which denominators are not same is called, unlike fractions.

e.g. 1/12, 5/7, 7/9 13/11 etc.

 

Compound Fraction: Fraction of a fraction is called a compound fraction.

e.g. 1/2 of 3/4 is a compound fraction.

 

Complex Fractions: Fractions in which numerator or denominator or both are fractions, are called complex fractions.

 

Continued fraction: Fraction that contain additional fraction is called continued fraction.

e.g.

 

 

 

Rule: To simplify a continued fraction, begin from the bottom and move upwards.

 

Decimal Fractions: Fractions in which denominators are 10 or multiples of 10 is called, decimal fractions. e.g. 1/10, 3/100, 2221/10000 etc.

 

Recurring Decimal: If in a decimal fraction a digit or a set of digits is repeated continuously, then such a number is called a recurring decimal. It is expressed by putting a dot or bar over the digits. e.g.

 

 

Pure recurring decimal: A decimal fraction in which all the figures after the decimal point is repeated is called a pure recurring decimal.

 

Mixed recurring decimal: A decimal fraction in which only some of the figures after the decimal point is repeated is called a mixed recurring decimal.

 

Conversion of recurring decimal into proper fraction: 

CASE I: Pure recurring decimal

 

Write the repeated digit only once in the numerator and put as many nines as in the denominator as the number of repeating figures. e.g.

 

CASE II: Mixed recurring decimal

In the numerator, take the difference between the number formed by all the digits after the decimal point and that formed by the digits which are not repeated. In the denominator, take the number formed as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. e.g.

 
Questions

Level-I

1.	Evaluate : (2.39)2 – (1.61)2 2.39 – 1.61
	A. 2 B. 4 C. 6 D. 8

 

2.	What decimal of an hour is a second ?
	A. .0025 B. .0256 C. .00027 D. .000126

 

3.	The value of (0.96)3 – (0.1)3 is: (0.96)2 + 0.096 + (0.1)2
	A. 0.86 B. 0.95 C. 0.97 D. 1.06

 

4.	The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02 is: 0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04
	A. 0.0125 B. 0.125 C. 0.25 D. 0.5

 

5.	If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
	A. 0.172 B. 1.72 C. 17.2 D. 172
6.	When 0.232323….. is converted into a fraction, then the result is:
	A. 1 5 B. 2 9 C. 23 99 D. 23 100

 

7.	.009 = .01 ?
	A. .0009 B. .09 C. .9 D. 9

 

8.	The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
	A. 0.02 B. 0.2 C. 0.04 D. 0.4

 

9.	(0.1667)(0.8333)(0.3333) is approximately equal to: (0.2222)(0.6667)(0.1250)
	A. 2 B. 2.40 C. 2.43 D. 2.50

 

10.	3889 + 12.952 – ? = 3854.002
	A. 47.095 B. 47.752 C. 47.932 D. 47.95
11.	Level-II     0.04 x 0.0162 is equal to:
	A. 6.48 x 10-3 B. 6.48 x 10-4 C. 6.48 x 10-5 D. 6.48 x 10-6

 

12.	4.2 x 4.2 – 1.9 x 1.9 is equal to: 2.3 x 6.1
	A. 0.5 B. 1.0 C. 20 D. 22

 

13.	If 144 = 14.4 , then the value of x is: 0.144 x
	A. 0.0144 B. 1.44 C. 14.4 D. 144

 

 

 

14.	The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
	A. 2010 B. 2011 C. 2012 D. 2013

 

15.	Which of the following are in descending order of their value ?
	A. 1 , 2 , 3 , 4 , 5 , 6 3 5 7 5 6 7 B. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 7 6 7 C. 1 , 2 , 3 , 4 , 5 , 6 3 5 5 6 7 7 D. 6 , 5 , 4 , 3 , 2 , 1 7 6 5 7 5 3
16.	Which of the following fractions is greater than 3 and less than 5 ? 4 6
	A. 1 2 B. 2 3 C. 4 5 D. 9 10

 

17.	The rational number for recurring decimal 0.125125…. is:
	A. 63 487 B. 119 993 C. 125 999 D. None of these

 

18.	617 + 6.017 + 0.617 + 6.0017 = ?
	A. 6.2963 B. 62.965 C. 629.6357 D. None of these

 

19.	The value of 489.1375 x 0.0483 x 1.956 is closest to: 0.0873 x 92.581 x 99.749
	A. 0.006 B. 0.06 C. 0.6 D. 6

 

20.	0.002 x 0.5 = ?
	A. 0.0001 B. 0.001 C. 0.01 D. 0.1

 

 

 

 

 

Answers

Level-I

Answer:1 Option B

 

Explanation:

Given Expression =	a2 – b2	=	(a + b)(a – b)	= (a + b) = (2.39 + 1.61) = 4.
	a – b		(a – b)

 

Answer:2 Option C

 

Explanation:

Required decimal =	1	=	1	= .00027
	60 x 60		3600

 

 

Answer:3 Option A

 

Explanation:

Given expression	= (0.96)3 – (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2
	= a3 – b3 a2 + ab + b2
	= (a – b)   = (0.96 – 0.1)   = 0.86

Answer:4 Option B

 

Explanation:

Given expression =	(0.1)3 + (0.02)3	=	1	= 0.125
	23 [(0.1)3 + (0.02)3]		8

 

 

 

 

Answer:5 Option C

 

Explanation:

29.94	=	299.4
1.45		14.5

 

=		2994	x	1		[ Here, Substitute 172 in the place of 2994/14.5 ]
		14.5		10

 

=	172
	10

= 17.2

 

 

Answer:6 Option C

 

Explanation:

0.232323… = 0.23 =	23
	99

 

Answer:7 Option C

 

Explanation:

Let	.009	= .01;     Then x =	.009	=	.9	= .9
	x		.01		1

 

 

Answer:8 Option C

 

Explanation:

Given expression = (11.98)² + (0.02)² + 11.98 x x.

For the given expression to be a perfect square, we must have

11.98 x x = 2 x 11.98 x 0.02 or x   = 0.04

 

Answer:9 Option D

 

Explanation:

Given expression	= (0.3333) x (0.1667)(0.8333) (0.2222) (0.6667)(0.1250)
	= 3333 x 1 x 5 6 6 2222 2 x 125 3 1000
	= 3 x 1 x 5 x 3 x 8 2 6 6 2
	= 5 2
	= 2.50

 

Answer:10 Option D

 

Explanation:

Let 3889 + 12.952 – x = 3854.002.

Then x = (3889 + 12.952) – 3854.002

= 3901.952 – 3854.002

= 47.95.

 

Level-II

Answer:11 Option B

 

Explanation:

4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10^-4

 

Answer:12 Option B

 

Explanation:

Given Expression =	(a2 – b2)	=	(a2 – b2)	= 1.
	(a + b)(a – b)		(a2 – b2)

 

 

Answer:13 Option A

 

Explanation:

144	=	14.4
0.144		x

 

	144 x 1000	=	14.4
	144		x

 

x =	14.4	= 0.0144
	1000

 

 

Answer:14 Option B

 

Explanation:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

z =	2.50	=	250	= 10.
	0.25		25

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

 

 

 

Answer:15 Option D

Answer:16 Option C

 

Explanation:

3	= 0.75,	5	= 0.833,	1	= 0.5,	2	= 0.66,	4	= 0.8,	9	= 0.9.
4		6		2		3		5		10

Clearly, 0.8 lies between 0.75 and 0.833.

	4	lies between	3	and	5	.
	5		4		6

 

 

 

Answer:17 Option C

 

Explanation:

0.125125… = 0.125 =	125
	999

 

 

Answer:18 Option C

 

Explanation:

617.00

6.017

0.617

+  6.0017

——–

629.6357

———

 

Answer:19 Option B

 

Explanation:

489.1375 x 0.0483 x 1.956		489 x 0.05 x 2
0.0873 x 92.581 x 99.749		0.09 x 93 x 100

 

=	489
	9 x 93 x 10

 

=	163	x	1
	279		10

 

=	0.58
	10

= 0.058  0.06.

 

Answer:20 Option B

 

Explanation:

2 x 5 = 10.

Sum of decimal places = 4

0.002 x 0.5 = 0.001